Solving Double Integrals: Finding J in R2

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squenshl
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How do I find
J := \int\int_U y cos(x + y) dA
where U = {(x,y) \in R2 : 0 \leq x \leq 2, 1-x \leq x \leq -2 + 2x}.

I drew the picture & i get 2 triangles but not sure what to do with them. Help please.
 
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in rectangular coordinates, dA=dx dy, so the region U describes will give you the limits for the integral.
 
I think you should only be getting one triangle. Try checking your picture against one generated by a graphing tool.

Cheers,
W. =)
 
Very true. I got a triangle & J = -0.5717341110 using maple.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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